In the given figure, `AB||CD` and `angleCDP=35^(@)`. PD is produced downwards to meet AB at E and `angleBEF=75^(@)`. If `DQ||EF`, find `angleAED, angleDEF` and `anglePDQ`.

Explanation

`AB||CD` and PDF is the transversal.
`:. angleAED=angleCDP=35^(@)` [corresponding `angles`].
Now, AB is a straight line and ray PDE stands on it.
`:. angleAED+angleDEF+angleBEF=180^(@)`
`implies 35^(@)+angleDEF+75^(@)=180^(@)`
`implies angleDEF=(180^(@)-110^(@))=70^(@)`.
Now, `DQ||EF` and PDE is the transversal.
`:. anglePDQ=angleDEF=70^(@)` [corresponding `angles`].
Thus, `angleAED=35^(@), angleDEF=70^(@)` and `anglePDQ=70^(@)`.